We say that the pair (u, v) of vectors is positively oriented if 0 &lt

The parallelogram spanned by the nonzero plane vectors u = (a, b) and v = (c, d) equals to the absolute value of

∆ =    

a c b d

    .

Assume that the angle from u to v is θ. We say that the pair (u, v) of vectors is positively oriented if 0 < θ < π and negatively oriented if π < θ < 2π.

Lemma 2.1. The pair of plane vectors (u, v) is positively (negatively) oriented if and only if ∆ > 0 (∆ < 0).

Proof. Let us use polar coordinate of R². Let r = |u| and ρ = |v| and φ be the angle from x-axis to u. Then u = r(cos φ, sin φ) and v = ρ(cos(φ + θ), sin(φ + θ)). Using the property of determinant, we see

a c b d

=    

r cos φ ρ cos(φ + θ) r sin φ ρ sin(φ + θ)

= rρ    

cos φ cos(φ + θ) sin φ sin(φ + θ)

= rρ (sin(φ + θ) cos φ − cos(φ + θ) sin φ) = rρ sin θ.

Since 0 < θ < π, sin θ > 0 and hence ∆ = rρ sin θ > 0.  Remark. When (u, v) is positively oriented, the area of the parallelogram spanned by u, v exactly equals ∆.

Definition 2.1. A subset Γ of R²is called a simple closed curve if there exists a function γ : [a, b] → R², t 7→ (x(t), y(t))

such that

(1) both x(t) and y(t) are continuous functions on [a, b], and (2) γ(a) = γ(b), and

(3) if γ(t) = γ(s), then t = s. (one-to-one), and (4) Γ = {γ(t) ∈ R²: t ∈ [a, b]}.

A simple closed curve is C¹if both x, y ∈ C¹[a, b].

Let Ω be a bounded plane region (a subset of R²) and Γ be its boundary, denoted by ∂Ω. Assume that Γ is a simple closed curve. Let us choose a point P in the interior of Ω and define θ : [a, b] → R be the angle between−−−→

P γ(t) and the horizontal line passing through P. Let us assume further that θ(b) = θ(a) + 2π and θ is a continuous function. We say that γ is positively oriented if θ is increasing on (a, b) and negatively oriented if θ is decreasing on (a, b).

1

Proposition 2.1. Let P0, · · · , Pn be n + 1-distinct points on R² with Pi = (xi, yi) and assume that these points determine a bounded plane region whose boundary is a simple closed polygon, positively oriented. Then the area of the polygon is given by

1 2

n+1

X

i=1

x_i−1 x_i yi−1 yi

    . Here we set P_n+1= P₀.

Now, let us assume that the boundary Γ of Ω is a simple closed C¹plane curve positively oriented.

Let P = {a = t0 < t1< · · · < tn = b} be a partition of [a, b]. Denote γ(ti) by Pi for 0 ≤ i ≤ n.

Notice that Pn= P0. Then the set of points {Pi: 1 ≤ i ≤ n} determines a polygon on R². We define the approximate area A(Ω, P) of Ω with respect to the partition P to be the area of the polygon defined by {Pi: 1 ≤ i ≤ n}.

Thus the approximate area of Ω with respect to P is given by A(Ω, P) = 1

2

n

X

i=1

x(ti−1) x(ti) y(ti−1) y(ti)    

Lemma 2.2. Let f, g be C¹-functions on [a, b]. Define a function h : [a, b] → R by h(x) =

f (a) f (x) g(a) g(x)

    .

Then h ∈ C¹[a, b]. Moreover, there exists c ∈ (a, b) such that h(b) = h⁰(c)(b − a). In other words, we

have 

f (a) f (b) g(a) g(b)

=    

f (a) f⁰(c) g(a) g⁰(c)

(b − a).

Proof. The proof follows from mean value theorem.  We can rewrite A(Ω, P) as

(2.1) A(Ω, P) = 1

2

n

X

i=1

x(ti−1) x⁰(ci) y(ti−1) y⁰(ci)    

(ti− ti−1)

for some c_i ∈ (ti−1, t_i). Unfortunately, the right hand side of (2.1) is not a Riemann sum. Let us rewrite (2.1) as follows:

A(Ω, P) = 1 2

n

X

i=1

x(ci) x⁰(ci) y(ci) y⁰(ci)    

(ti− ti−1) (2.2)

+1 2

n

X

i=1

   

x(t_i−1) x⁰(c_i) y(t_i−1) y⁰(c_i)

−    

x(c_i) x⁰(c_i) y(c_i) y⁰(c_i)    



(ti− ti−1) (2.3)

Let us define F : [a, b] → R by

F (t) =    

x(t) x⁰(t) y(t) y⁰(t)

    .

Since x, y ∈ C¹[a, b], F ∈ C[a, b]. The right hand side of (2.2) is a Riemann sumPn

i=1F (c_i)(t_i−ti−1) of F (t) with respect to the partition P of mark C = {c_i}. By the continuity of F, this Riemann sum is convergent to the integral of F (t) over [a, b] which is

I = Z b

a

x(t) x⁰(t) y(t) y⁰(t)    

dt.

Lemma 2.3. Let f, g be C¹ functions on [a, b]. Define a function h : [a, b] → R by h(x) =

f (x) f⁰(b) g(x) g⁰(b)     .

For any x₁, x₂∈ [a, b], there exists c ∈ (x1, x₂) such that h(x₂) − h(x₁) = h⁰(c)(x₂− x1). In other words, we have

f (x2) − f (x1) f⁰(b) g(x₂) − g(x₁) g⁰(b)    

=    

f⁰(c) f⁰(b) g⁰(c) g⁰(b)     .

Proof. By mean value theorem. 

Using this lemma, we obtain that 

x(ti−1) x⁰(ci) y(ti−1) y⁰(ci)    

−    

x(ci) x⁰(ci) y(ci) y⁰(ci)    

=    

x⁰(di) x⁰(ci) y⁰(di) y⁰(ci)    

(ci− ti−1).

By continuity of x⁰, y⁰on [a, b] and the Weierstrass theorem, x⁰, y⁰are bounded functions on [a, b]. As- sume that |x⁰(t)|, |y⁰(t)| are both bounded by M on [a, b]. Thus the absolute value of the determinant 

x⁰(di) x⁰(ci) y⁰(di) y⁰(ci)

is bounded by 2M²:

|x⁰(d_i)y⁰(c_i) − y⁰(d_i)x⁰(c_i)| ≤ |x⁰(d_i)||y⁰(c_i)| + |y⁰(d_i)||x⁰(c_i)| ≤ M · M + M · M = 2M². Moreover, |ci− ti−1| < ti− ti−1 ≤ kP k for all i. Thus the absolute value of (2.3) is bounded by M²kP k(b − a) :

     1 2

n

X

i=1

   

x(t_i−1) x⁰(c_i) y(t_i−1) y⁰(c_i)    

−    

x(c_i) x⁰(c_i) y(c_i) y⁰(c_i)    



(ti− ti−1)     

≤1 2

n

X

i=1

2M²kP k(ti− ti−1)

= M²kP k

n

X

i=1

(t_i− ti−1)

= M²kP k(b − a).

Theorem 2.1. Let Ω be a plane region. Suppose its boundary Γ = ∂Ω is a simple closed C¹-curve with positive orientation. For every  > 0, there exists δ> 0 such that

|A(Ω, P) − I| <  whenever kP k < δ. Proof. By the continuity of F, given  > 0, we may choose δ⁰_> 0 such that

n

X

i=1

F (ci)(ti− ti−1) − I     

<  2. Let δ= min{δ_⁰, /2M²(b − a)}. Therefore

     1 2

n

X

i=1

   

x(t_i−1) x⁰(c_i) y(t_i−1) y⁰(c_i)    

−    

x(c_i) x⁰(c_i) y(c_i) y⁰(c_i)



(t_i− ti−1)     

<  2. Using the equality

A(Ω, P)−I =

n

X

i=1

F (ci)(ti− ti−1) − I

! +1

2

n

X

i=1

   

x(t_i−1) x⁰(c_i) y(t_i−1) y⁰(c_i)    

−    

x(c_i) x⁰(c_i) y(c_i) y⁰(c_i)    



(ti−ti−1) and the triangle inequality, we find

|A(Ω, P) − I| ≤  2+ 

2 = .

We prove our assertion. 

Definition 2.2. Let Ω and Γ be as above. Then the area of Ω is defined to be A(Ω) = 1

2 Z b

a

x(t) x⁰(t) y(t) y⁰(t)

dt = 1 2

Z b a

(x(t)y⁰(t) − y(t)x⁰(t))dt.

Using integration by parts, we also obtain A(Ω) =

Z b a

x(t)y⁰(t)dt = − Z b

a

y(t)x⁰(t)dt.

Suppose P (x, y) and Q(x, y) are functions in x, y. Let Γ be a C¹-parametrized curve with parametrization x = x(t), y = y(t) for a ≤ t ≤ b. Assume that P (x(t), y(t)) and Q(x(t), y(t)) are continuous functions on [a, b]. We define

Z

Γ

P (x, y)dx + Q(x, y)dy = Z b

a

{P (x(t), y(t))x⁰(t) + Q(x(t), y(t))y⁰(t)}dt.

Thus the area formula can be rewritten as A(Ω) = 1

2 Z

Γ

(xdy − ydx) = Z

Γ

xdy = − Z

Γ

ydx.

Example 2.1. Verify that the area of the closed disk D = x²+ y²≤ a² is πa². Solution:

The boundary of the closed disk D is the circle C^a : x²+ y²= a². We know Ca is a simple closed curve. We take

x = a cos t, y = a sin t, for 0 ≤ t ≤ 2π.

Then Ca is parametrized and positively oriented. We compute dx = −a sin tdt, dy = a cos tdt.

Hence

xdy − ydx = {a cos t · a cos t − a sin t · (−a sin t)}dt = a²dt.

Thus the area of D is given by

A(D) = 1 2

Z 2π 0

a²dt = πa².

Example 2.2. Let Ω be a plane region. Suppose Γ is simple, closed C¹ and positively oriented.

Suppose r = r(θ), α ≤ θ ≤ β is the parametrization of Γ using polar coordinate system of R². Show that

A(Ω) = 1 2

Z β α

r(θ)²dθ.

Proof. We know x(θ) = r(θ) cos θ and y(θ) = r(θ) sin θ. Then

x⁰(θ) = r⁰(θ) cos θ − r(θ) sin θ, y⁰(θ) = r⁰(θ) sin θ + r(θ) cos θ.

Hence

x(θ)y⁰(θ) − y(θ)x⁰(θ) = r(θ) cos θ(r⁰(θ) sin θ + r(θ) cos θ) − r(θ) sin θ(r⁰(θ) cos θ − r(θ) sin θ)

= r(θ)²(cos²θ + sin²θ) = r(θ)². The area of Ω is given by

A(Ω) = 1 2

Z β α

r(θ)²dθ.